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3 years ago

May you please help?

Physics

1 answer:

kirza4 [7]3 years ago

3 0

Choice-A is the main reason that people use the thing in the picture.

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A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

How are the electric field lines around a positive charge affected when a second positive charge is near it? The field lines com

timofeeve [1]

The answer would be D hope it helps and sorry if it is wrong. :)

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4 years ago

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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m

Irina18 [472]

The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.

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3 years ago

The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter

ryzh [129]

Answer:

Explanation:

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3 years ago

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In 2h2so4, the number of hydrogen atoms is 1 2 4 8

pishuonlain [190]

C.) In this, number of Hydrogen atoms is 4

7 0

3 years ago

May you please help?​

May you please help?